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05A05-RuleOfProduct.tex
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65 lines (53 loc) · 2.34 KB
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\documentclass[12pt]{article}
\usepackage{pmmeta}
\pmcanonicalname{RuleOfProduct}
\pmcreated{2013-03-22 19:13:02}
\pmmodified{2013-03-22 19:13:02}
\pmowner{pahio}{2872}
\pmmodifier{pahio}{2872}
\pmtitle{rule of product}
\pmrecord{6}{42137}
\pmprivacy{1}
\pmauthor{pahio}{2872}
\pmtype{Definition}
\pmcomment{trigger rebuild}
\pmclassification{msc}{05A05}
\pmclassification{msc}{03-00}
\pmsynonym{multiplication principle}{RuleOfProduct}
\pmrelated{CartesianProduct}
\pmrelated{Combinatorics}
\pmrelated{Cardinality}
\pmrelated{Number}
\pmrelated{Product}
\endmetadata
% this is the default PlanetMath preamble. as your knowledge
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\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
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%\usepackage{psfrag}
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%\usepackage{graphicx}
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\usepackage{amsthm}
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%%%\usepackage{xypic}
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\theoremstyle{definition}
\newtheorem*{thmplain}{Theorem}
\begin{document}
\PMlinkescapeword{combination}
If a process $A$ can have altogether $m$ different results and another process $B$ altogether $n$ different results, then the two processes can have altogether $mn$ different combined results.\, Putting it to set-theoretical form,
$$\mbox{card}(A\!\times\!B) \;=\; m\!\cdot\!n.$$
The \emph{rule of product} is true also for the combination of several processes:\, If the processes $A_i$ can have $n_i$ possible results ($i = 1,\,2,\,\ldots,\,k$), then their combined process has $n_1n_2\!\cdots\!n_k$ possible results.\, I.e.,
$$\mbox{card}(A_1\!\times\!A_2\!\times\ldots\times\!A_k) \;=\; n_1n_2\!\cdots\!n_k.$$\\
\textbf{Example.}\, Arranging $n$ elements, the first one may be chosen freely from all the $n$ elements, the second from the remaining $n\!-\!1$ elements, the third from the remaining $n\!-\!2$, and so on, the penultimate one from two elements and the last one from the only remaining element; thus by the rule of product, there are in all
$$n(n\!-\!1)(n\!-\!2)\!\cdots\!2\!\cdot\!1 \;=\; n!$$
different arrangements, i.e. permutations, as the result.
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\end{document}